Answer to Question #311844 in Chemistry for grey tyrant

Question

Lead(II) sulfate formed in the reaction of 145 mL of 0.123 M lead(II) nitrate and excess sodium sulfate.

c. Calculate the mass of lead(II) sulfate formed in the reaction

Answer 1

Solution:

lead(II) nitrate = Pb(NO₃)₂

sodium sulfate = Na₂SO₄

lead(II) sulfate = PbSO₄

Calculate the moles of Pb(NO₃)₂:

Moles of Pb(NO₃)₂ = Molarity of Pb(NO₃)₂ × Volume of Pb(NO₃)₂

Moles of Pb(NO₃)₂ = (0.123 M) × (0.145 L) = 0.017835 mol

Balanced chemical equation:

Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2NaNO₃

According to stoichiometry:

1 mol of Pb(NO₃)₂ produces 1 mol of PbSO₄

Thus, 0.017835 ol of Pb(NO₃)₂ produces:

[0.017835 mol Pb(NO₃)₂] × [1 mol PbSO₄ / 1 mol Pb(NO₃)₂] = 0.017835 mol PbSO₄

Calculate the mass of PbSO₄:

The molar mass of PbSO₄ is 303.26 g/mol

Therefore,

Mass of PbSO₄ = (0.017835 mol PbSO₄) × (303.26 g PbSO₄ / 1 mol PbSO₄) = 5.41 g PbSO₄

Mass of PbSO₄ = 5.41 g PbSO₄

Answer: 5.41 grams of lead(II) sulfate (PbSO₄) formed in the reaction